Problem 102 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the logarithmic equation algebraically. Round the result to three decimal places. Verify your answer(s) using a graphing utility. $$4 \log _{10}(x-6)=11$$

Step-by-Step Solution

Answer

The solution to the logarithmic equation \(4 \log _{10}(x-6)=11\) rounded off to three decimal places is \(x = 70.853\).

Step 1: Rearrange The Equation

First step is to isolate the logarithm on one side of the equal sign: \[4\log_{10}(x-6) = 11 \implies \log_{10}(x-6) = \frac{11}{4}\]

Step 2: Convert Log Form to Exponential Form

Once the equation is isolated, we can now convert it to exponential form using the property of logs \( \log_b a = c \rightarrow a = b^c \). It looks like this: \[10^\frac{11}{4} = x - 6\]

Step 3: Solving for x

Now, solving for x involves adding 6 to both sides of the equation: \[(10^\frac{11}{4}) + 6 = x\]

Step 4: Rounding the solution

To solve for x and ensure it’s accurate to three decimal places, the resulting number is rounded: \[(10^\frac{11}{4}) + 6\] will give \(x =70.853\).

Key Concepts

Properties of Logarithms

Understanding the properties of logarithms is essential when solving logarithmic equations. Logarithms have a few key properties that make manipulating equations much easier. For example, if you have a logarithm with multiplication inside, you can break it into a sum of logs: \(\log_b(m \cdot n) = \log_b m + \log_b n\). If you have a division inside the log, it becomes a difference: \(\log_b\left(\frac{m}{n}\right) = \log_b m - \log_b n\). Another crucial property involves the power rule: \(\log_b(m^n) = n \log_b m\).
These properties help streamline complex expressions, making it easier to solve equations algebraically. Understanding them equips you to convert logarithmic equations and efficiently solve for unknowns.

Solving Equations Algebraically

When faced with logarithmic equations, solving them algebraically involves isolating the logarithmic part and manipulating the equation to find the solution. The general steps include:

• First, rearrange the equation to isolate the logarithmic term on one side, as seen in the example where \(4\log_{10}(x-6)=11\) becomes \(\log_{10}(x-6)=\frac{11}{4}\).
• Use properties of logarithms, if necessary, to simplify the expression further.
• Prepare to convert to an exponential equation, which often makes it easier to solve.

By methodically isolating the log and then using algebraic manipulation, you can solve the equation for the variable. This approach ensures that you follow a clear path to find the answer.

Graphical Verification

Graphical verification serves as an excellent way to confirm algebraic solutions. Once an equation like \(4\log_{10}(x-6)=11\) has been solved algebraically, graphing utility can be used to verify the solution. Here's how you can do it:
Separate functions for each side of the equation:

• \(f(x) = 4\log_{10}(x-6)\)
• \(g(x) = 11\)

Plot both functions on the same graph. The intersection point on the graph where \(f(x) = g(x)\) should match your algebraic solution for \(x\).
This method visually confirms the solution, providing a check against errors. Moreover, it can help you understand the behavior of the function across different values of \(x\).

Exponential Form

Logarithmic and exponential forms are closely related, making them interchangeable with the right manipulations. To turn a logarithmic equation into exponential form, remember the relationship: if \(\log_b a = c\), then \(a = b^c\). This transformation is key to solving a logarithmic equation algebraically.
In our example, after isolating the logarithm so that \(\log_{10}(x-6) = \frac{11}{4}\), it is converted into exponential form as \(10^{\frac{11}{4}} = x-6\). This step takes us to a form where we can easily solve for \(x\), allowing completion by simple arithmetic operations.
By changing the equation into an exponential form, you simplify complex calculations and hone in on the solution efficiently. This technique underscores the flexibility and power of understanding the links between exponential functions and logarithms.